Taboos

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Thanks for the patient explanation Prebe. Most interesting. (I'll frame the diploma.)

Now that we've established that, in any isolated incident in an open population, incest has only a remote chance of producing a gentically "damaged" offspring (if one were to be produced at all), the only remaining factors are the psychological and social consequences.

--A

[Prebe]

The likelihood is still larger than when the mating individuals are unrelated, as 25% of all loci (in the most extreme inbreeding parent-child) will be homozygous IBD (have two identical alleles at one in four loci identical by descent).

Simply: The likelihood of suffering from ANY recessive genetical dissease, depends on how many loci that are homozygous (having two identical alleles). Any inbreeding (regardles of how often it goes on) increases the number of homozygous loci.

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More than fair enough...an increased likelihood is a given obviously. And any already present recessive would have a significantly higher chance of being reinforced because of the the presence of homozygous alleles, compared to unrelated breeding which would at least guarantee heterozygosity.

So any single offspring is more likely to exhibit a defect, even within that single generation? (Assuming a negative reccesive is not only present, but part of that 25% homozygosity.) So even though it wouldn't affect the population per se, it would still have a (high?) chance of affecting the offspring? Makes sense...I suppose that the chance of actual effect fluctuates depending on the existing genetic make-up (i.e. how many bad alleles) of the people involved?

Still, an unacceptably high chance either way. Not that anybody thinks it's a good idea to have children by relatives. If I recall the original question correctly, it specified that no offspring were possible, in which case it is merely a matter of social/cultural mores and taboo's.

--A

[Prebe]

So any single offspring is more likely to exhibit a defect, even within that single generation? (Assuming a negative reccesive is not only present, but part of that 25% homozygosity.)

Yes. Example: Dad has an incredibly rare allele, that he passes on to his daughter (at a 50% chance). He then goes on to impregnate his daughter (Loth spings to mind). The child of that union has a 50% of getting the allele from dad, and a 50% chance of getting it from mum (who got it from dad).

So the likelihood that a child of a parent/child relation will exhibit the trait (have the rare allel in two copies at that locus) is 0.5^3 = 12.5% even with no previous inbreeding. Notice that this is half the likelihood of being homozygous IBD at the locus, because there is the same chance that the homozygosity is the "safe" allele.

If you multiply this likelihood with the likelihood that dad possesses one or more rare dissease recessive alleles, you will have the liklihood, that the ofspring exhibits any negative recessive trait.

P.S. This is something I can discuss with a clean conscience while at work, since it is kinda.... my work

[Avatar]

Is there a figure for the likelihood of any given person possessing one or more recessive rare disease alleles?

And would that multiplication be right? Surely this example reflects the chance of that recessive being present in the parent as 100%? I.e. if there is such a recessive present, then there is a 12.5% chance of it being passed to the offspring of an interbred couple?

So then surely you can't multiply the percentages can you? Afterall, if there is a 10% chance of the presence of a recessive, and a 12.5% chance of negative reinforcement, then multiplyig those would give you 125% wouldn't it? (Unless there's something I'm neglecting about multiplying pecentages...very possible...math is not my forte. )

--A

[I'm Murrin]

10/100 * 12.5/100 = 1.25%
Fractions. They get smaller when you multiply.

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Thanks Murrin...suspected I was doing something wrong.

--A

[Prebe]

So then surely you can't multiply the percentages can you? Afterall, if there is a 10% chance of the presence of a recessive, and a 12.5% chance of negative reinforcement, then multiplyig those would give you 125% wouldn't it? (Unless there's something I'm neglecting about multiplying pecentages...very possible...math is not my forte. )

There is something you are neglecting. Percentage is fractions of 100. So 10% = 0.1 and 12.5% = 0.125
The result would be 0,0125 or 1.25%.

As long as events are independent you can alway multiply their likelihood to find out the combined probability. E.g. The chances of rain tomorrow is 10% and the chance that the special of today in the cafeteria is Fish'n Chips is 50%. The combined probability (that it rains tomorrow AND I'll have F&C) is 5%.

In our example the liklihood that the father carries a deleterious recessive allele is dependent on the allele frequencies in the population, and the likelihood that the inbred ofspring becomes a homozygous for any given bad recessive allele depends on the way the cromosomes combine durring mating. So the calculation is perfectly in order.

Edit: Thanks Murrin

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Thanks, you may make fun of my mathematical ineptitude if you like.

(Actually, I did convert the percentages to decimals and multiply them, but when I got 0.0125 I was further confused, and couldn't turn that back into a percentage. )

So is there is figure for the likelihood of carrying a negative recessive? Or is it impossible to generalise that?

--A

[Prebe]

Thanks, you may make fun of my mathematical ineptitude if you like.

Not laughing at you Av, laughing right next to you

So is there is figure for the likelihood of carrying a negative recessive? Or is it impossible to generalise that?

Nobody is going to be able to give you that answer. But it can be calculated, only if you know the allele frequencies of all deleterious recessives in your population. If you have say three alleles (frequencies p, q and r) the probability that any person carries at least one of these is:

P(at least one bad allele) = 1-((1-p)*(1-q)*(1-r))

P(X) denotes the probability of X.

p, q and r are very small numbers. It should be evident from the formula (heh!), that the probability is one minus the product of probabilities that a person does NOT have allele p, q or r. It should also be evident that the more alleles you look at, the smaller the number that you must subtract from one is going to be hence the more likely the person is to have at least one of them.

Your insistance on understanding this difficult subject is impressive!

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I just like to know how things work. The more things I know, the more I like it. (And the more I realise how little I know. )

Thanks again. So while it's possible, it's effectively impossible to predict for any random person, simply because we wouldn't necessarily know the frequencies of all the bad recessives he could be carrying.

Still, considering that even with a 100% certainty of a bad recessive, there's still only a 12.5% chance of negative reinforcement, statistically, it's not that dangerous randomly across a single generation. But clearly not a good idea for the species.

Fascinating subject really.

--A